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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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The "mysterious" John Karl's book formulas
From: Antonio Sauci
Date: 2018 Jan 18, 21:28 -0800

To the navigational triangle,apply Bessel group II formula:
cos(90-H) = cos(90-d)xcos(90- latitude) + sin(90-d) x sin(90-latitude) x cos(LHA)               (1)
Let col = 90- latitude                                                                                                            (2)
Then (1) becomes:
sin(H) = sin(d) x cos(col) + cos(d) x sin(col) x cos (LHA)                                                     (3)
Let cot(G) = cot(d) x cos(LHA)                                                                                             (4)
By putting (4) into (3) we get
sin(H) = sin(d)[cos(col) + sin(col) x (cos(G)/sin(G))] = (sin(d)/sin(G)) x sin (G + col)
whence
sin(G+col) = sin(H) x sin(G) / sin(d)                                                                                      (5)
Eqn. (4) can be rewritten as
1/tan(G) = (1/tan(d)) x cos(LHA), or
tan(G) = tan(d) / cos(LHA)                                                                                                   (6)

I hope that my two cents helps Chris Caswell´s question. Browse Files

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